0
$\begingroup$

What I am taught is that electric field is the direction in which force is experienced by the positive charge particle and the amount of force is determined by the density of the electric field. If two positive charge are kept d distance apart then how will the electric field line behave which emerges from the point O and P on the charge particles?

enter image description here

what I think is that if we place a positive test charged particle near O(initial velocity=0) then test particle will experience a force pushing it away from the charge particle and finally the particle will reach the equilibrium point where there will no force. the field line must be the line tangent to all the force vectors which acted upon the test particle. No where the direction of the force acting on the particle is changed that means the force line must be straight emerging from the charge particle and since at the equilibrium point there must be no force hence no field line. In every textbook the field line is curved upwards but is is clear that no where the test charge had experience a force with upward component. So how will the force line will proceed?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

From mathematical perspective, your perfect field line (called separatrix) will end in a saddle point (see the picture below). However, physics doesn't talk a lot about separatrices, because any small deviation will bump the test particle either in upper or in lower half-plane.

enter image description here

$\endgroup$
2
  • $\begingroup$ What does the R1, R2 and I1, I2 stands for? $\endgroup$ Mar 26, 2020 at 23:44
  • $\begingroup$ The stuff under the image is likely not important to the image itself $\endgroup$
    – Triatticus
    Mar 27, 2020 at 6:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.